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3 years ago

If cos(-ϴ)=0.95, find sin(π/2-ϴ)

1 answer:

6 0

$\bf \textit{Cofunction Identities}
\\\\
sin\left(\frac{\pi}{2}-\theta\right)=cos(\theta)
\qquad 
cos\left(\frac{\pi}{2}-\theta\right)=sin(\theta)
\\\\\\
\stackrel{\textit{symmetry identities}}{cos(-\theta )=cos(\theta )}\\\\
-------------------------------\\\\
sin\left(\frac{\pi}{2}-\theta\right)=cos(\theta)=cos(-\theta )=0.95$

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Check whether 7+3x is a factor of 3x^3+7x

BartSMP [9]

Answer:

Step-by-step explanation:

refer to picture for solution

3 0

3 years ago

find the angle between the vectors. (first find the exact expression and then approximate to the nearest degree. ) a=[1,2,-2]. B

SashulF [63]

Answer:

$\theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190$

Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel

And the anfle is approximately $\theta \approx 48$

Step-by-step explanation:

For this case first we need to calculate the dot product of the vectors, and after this if the dot product is not equal to 0 we can calculate the angle between the two vectors in order to see if there are parallel or not.

a=[1,2,-2], b=[4,0,-3,]

The dot product on this case is:

$a b= (1)*(4) + (2)*(0)+ (-2)*(-3)=10$

Since the dot product is not equal to zero then the two vectors are not orthogonal.

Now we can calculate the magnitude of each vector like this:

$|a|= \sqrt{(1)^2 +(2)^2 +(-2)^2}=\sqrt{9} =3$

$|b| =\sqrt{(4)^2 +(0)^2 +(-3)^2}=\sqrt{25}= 5$

And finally we can calculate the angle between the vectors like this:

$cos \theta = \frac{ab}{|a| |b|}$

And the angle is given by:

$\theta = cos^{-1} (\frac{ab}{|a| |b|})$

If we replace we got:

$\theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190$

Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel

And the anfle is approximately $\theta \approx 48$

3 0

3 years ago

A simple random sample of size nequals10 is obtained from a population with muequals68 and sigmaequals15. (a) What must be true

valentina_108 [34]

Answer:

(a) The distribution of the sample mean ( $\bar x$ ) is N (68, 4.74²).

(b) The value of $P(\bar X$ is 0.7642.

Step-by-step explanation:

A random sample of size n = 10 is selected from a population.

Let the population be made up of the random variable X.

The mean and standard deviation of X are:

$\mu=68\\\sigma=15$

(a)

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Since the sample selected is not large, i.e. n = 10 < 30, for the distribution of the sample mean will be approximately normally distributed, the population from which the sample is selected must be normally distributed.

Then, the mean of the distribution of the sample mean is given by,

$\mu_{\bar x}=\mu=68$

And the standard deviation of the distribution of the sample mean is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{10}}=4.74$

Thus, the distribution of the sample mean ( $\bar x$ ) is N (68, 4.74²).

(b)

Compute the value of $P(\bar X$ as follows:

$P(\bar X$

$=P(Z$

*Use a z-table for the probability.

Thus, the value of $P(\bar X$ is 0.7642.

(c)

Compute the value of $P(\bar X\geq 69.1)$ as follows:

Apply continuity correction as follows:

$P(\bar X\geq 69.1)=P(\bar X> 69.1+0.5)$

$=P(\bar X>69.6)$

$=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{69.6-68}{4.74})$

$=P(Z>0.34)\\=1-P(Z$

Thus, the value of $P(\bar X\geq 69.1)$ is 0.3670.

7 0

3 years ago

The number of major earthquakes in a year is approximately normally distributed with a mean of 20.8 and a standard deviation of

SCORPION-xisa [38]

Answer:

a) 51.60% probability that in a given year there will be less than 21 earthquakes.

b) 49.35% probability that in a given year there will be between 18 and 23 earthquakes.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 20.8, \sigma = 4.5$

a) Find the probability that in a given year there will be less than 21 earthquakes.

This is the pvalue of Z when X = 21. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{21 - 20.8}{4.5}$

$Z = 0.04$

$Z = 0.04$ has a pvalue of 0.5160.

So there is a 51.60% probability that in a given year there will be less than 21 earthquakes.

b) Find the probability that in a given year there will be between 18 and 23 earthquakes.

This is the pvalue of Z when X = 23 subtracted by the pvalue of Z when X = 18. So:

X = 23

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{23 - 20.8}{4.5}$

$Z = 0.71$

$Z = 0.71$ has a pvalue of 0.7611

X = 18

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{18 - 20.8}{4.5}$

$Z = -0.62$

$Z = -0.62$ has a pvalue of 0.2676

So there is a 0.7611 - 0.2676 = 0.4935 = 49.35% probability that in a given year there will be between 18 and 23 earthquakes.

5 0

3 years ago

Please help me, pic provide d

vfiekz [6]

Fill in the y of the bottom equation with the top equation because it gives you what Y equals. To answer the question it would be easier to input the top equation into the bottom one. Once you then get the equation for X go to the equation on top and fill in the x to get the y. X=0 Y=2

5 0

3 years ago